Fall 2015 Math 151
Final Exam Practice - Answers courtesy:
Amy Austin
7. a.) 3
b.) −2
c.) −5
d.) The limit does not exist
Final Exam Practice: Sections 1.1 - 6.4
e.) Not continuous at x = 3 (not in domain), not continuous at x = −1, x = 5 and x = 7 (the limit does not
exist). Not differentiable at x = −1, x = 3, x = 5 and
x = 7 (not continuous implies not differentiable). Also
not differentiable at x = −4 and x = −6 because of
sharp corners.
1. a.) h−5, −7i
−1 −3
b.) √ , √
10
10
◦
c.) 153
d.) Vector projection:
2 6
,
;
5 5
−4
scalar projection: √
10
8. f ′ (x) =
2. A vector equation: h1 + 2t, −2 + 10ti;
x→1
f (1) = 4.
100
√ √ N,
(1 + 3) 2
100
√ N
Magnitude of below force:
1+ 3
c.) a = 6, b = −3
4. Magnitude of above force:
f (x + h) − f (x)
h→0
h
5. Formula to use: f ′ (x) = lim
1
a.) f ′ (x) = √
2 1+x
−1
b.) f ′ (x) =
(x − 3)2
10. x + y + 1 = 0, 11x − y = 25
11. horizontal asymptote: y = 0, vertical asymptote: x = 1.
12.
5
12
13. 4
14. x > ln 4
6. a.) ∞
1
b.)
4
1
c.) −
4
d.) The limit does not exist because lim f (x) = 17
15. f −1 (x) = ln
16. a.) y ′ =
b.) y ′ =
x→3+
e.) −3
1
f.)
5
1
g.) −
5
h.) DNE
1
i.) −
2
1
j.)
2
if x < 0 or x > 2
if 0 < x < 2
b.) Not continuous at x = 1 because lim f (x) = 1, yet
3. x = 8 + 4t, y = 5 + 9t
x→3
2x − 2
−2x + 2
9. a.) Not continuous at x = 0 because lim f (x) does not
x→0
exist. Continuous for all other values of x.
parametric equations: x = 1 + 2t, y = −2 + 10t
and lim− f (x) = 5
x
1−x
1 − 3x2 y + 9x2
x3 + 4y 3 − 1
sin(x − y) + 2y − 4
sin(x − y) − 2x
17. a.) f ′ (x) =
12x2 − 4x4 − 16x
(1 − x2 )2
b.) f ′ (t) = 3t2 cos(1 − t2 ) + 2t4 sin(1 − t2 )
c.) G′ (x) = 12 tan2 (4x − 1) sec2 (4x − 1)
18. 56
19. y − ln 27 = (3 + ln 27)(x − ln 3)
20. At t = −1: y + 17 = 6(x + 1); horizontal tangent:
y = 64; verical tangent: x = 0
21. L(x) = ln 2 + 1/2(x − 2),
1
1
Q(x) = ln 2 + (x − 2) − (x − 2)2 . The linear and
2
8
quadratic approximations are useful in approximating
the function for x sufficiently close to a.
22.
49
dh
=
cm/min
dt
36π
39.
40. lim
n 1 P
n→∞ i=1
n
(3 + i/n)2 + 5(3 + i/n)
√
41. 5/ 29
42.
23. 5/3 feet per second
1 1/64
e
+ e9/64 + e25/64 + e49/64
4
π
3
24. 0.5 feet per second
3
1
43. y ′ = − √
−√
t(1 + t)
1 − 9t2
25. x = 2
44. 1,
2e
Since this is not in the domain, there is no
1−e
solution.
26. x =
27. t =
2
5
3
2
√
90
45. 2 30 by √
30
46. 3x6x2 cubic feet
x2
+3
2
28. x = t, y = e + 2et
47. f (x) = − cos x + 5ex +
29. f ′ (x) = − tan x
48. h(t) = −4.9t2 − 5t + 350
√
49. 2x 1 − x8
30.
y′
=y
!
√
ln(1 + x + 3x3 )
x(1 + 9x2 )
√
+
, where
2 x
1 + x + 3x3
√
y = (1 + x + 3x3 )
50. 5
x
51. π
31. 7.284 minutes
52.
32. t = .944 hours
33. Inc: (3, ∞), Dec: (−∞, 3), Local Min: (3, −33); Local
Max: None; Concave up: (−∞, 0) and (2, ∞), concave
down: (0, 2), points of inflection: (0, −6) and (2, −22)
34. (0, ∞)
35. Absolute Max: -1; Absolute min: -5
36. critical values: x = −1, x = 1, x = 5, f inc:
(−1, 1), (5, ∞), ; f dec: (−∞, −1), (1, 5); local min:
x = −1,x = 5; local max: x = 1; f cu: (−∞, 0) and
(5, ∞); f cd: (0, 4); inflection points: x = 0, x = 5.
37. a.) −3
b.) e3
c.) 7
d.) 0
−2
e.)
π
38.
2 5/2 4 3/2
x − x + 2x1/2 + C
5
3
140
− 4 ln 4
3
3π
4
π
54.
6
53.